Question: The value $2^8 -1$ is divisible by 3 prime numbers. What is the sum of the three prime numbers?
Factor twice using difference squares to obtain $(2^8-1)=(2^4+1)(2^4-1)=(2^4+1)(2^2+1)(2^2-1)=17\cdot5\cdot3$. The sum of the 3 prime factors of $2^8-1$ is $17+5+3=\boxed{25}$.